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Floats on Mercury

  1. Jun 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A 17.89 cm diameter, 42.43 cm tall steel cylinder (rsteel=7900 kg/m3) floats in mercury. The axis of the cylinder is perpendicular to the surface. What length of steel is above the surface?


    2. Relevant equations

    Pb=Pa+density*g*h


    3. The attempt at a solution

    Pa=101300
    Pb?
    density= 7900 kg/m^3
    H=(total length-submerged)
    g=(9.8)

    Pb=101300+((9.8*7900*(total length-submerged))

    I have 2 unkowns. What would Pb be?
     
  2. jcsd
  3. Jun 11, 2009 #2
    Any help would be much apperiated!
     
  4. Jun 11, 2009 #3
    For some reason I am having a hard time grasping pressure.
     
  5. Jun 11, 2009 #4

    Redbelly98

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    It looks like you're using Pb for the pressure on the bottom face. You can express that in terms of the density of mercury and the depth.
     
  6. Jun 11, 2009 #5
    Will Pb(pressure on bottom) always equal the denisty * depth? That is the density of the force acting against the object in this case......mercury?
     
  7. Jun 11, 2009 #6
    so i have RHO(mercury)*H(submerged)=101300(Pa)+(RHO*G*LENGTH OF CYLINDER)

    solve for h i have (101300+(7900*9.8*.4243))/13600= 9.8 m.....?
     
  8. Jun 11, 2009 #7
    i tried 0 seeing if it was a trick question but the trick was on me because it was wrong
     
  9. Jun 11, 2009 #8

    Redbelly98

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    Almost. It's actually ρmercury·g·hsubmerged, plus the pressure at the surface of the mercury.
     
  10. Jun 11, 2009 #9
    So Pb= pressure on surface of mercury+denisty *g*h? that doesn't seem right
     
  11. Jun 12, 2009 #10

    Redbelly98

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    What about it does not seem right to you?

    Things about it that are right (if you think about it):

    • Pressure at the surface (h=0) is the pressure at the surface, i.e. atmospheric pressure.
    • Pressure increases with depth h
    • Pressure is greater for denser fluids
    • Pressure is greater for greater values of g
     
  12. Jun 12, 2009 #11
    What doesn't seem right is that Pb is pressure at bottom, so why would i need the pressure at the top.
    Pb=Pa+rho*g*h
    ^ ^
    ^ ^
    ^ Pressure at above which in this case is atmosphereic pressure
    ^
    Pressure at bottom

    So if the pressure at the bottom is just (density of mercury*h*g) that works right? Where h is the amount submerged

    The pressure at the top/above is jsut 101300 Pa

    rho is denisty of the stell which is given 7900, g is 9.8 and h is the total height

    Where am I going wrong?
     
  13. Jun 12, 2009 #12

    Redbelly98

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    No, that doesn't work. What happens when h is zero? According to you, the pressure would be zero. But we know that the pressure actually equals atmospheric pressure at h=0.
     
  14. Jun 12, 2009 #13
    Isn't that why we have pressure above the can?
     
  15. Jun 12, 2009 #14
    Is there another way to appoarch this?
     
  16. Jun 12, 2009 #15

    Redbelly98

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    Last edited by a moderator: Apr 24, 2017
  17. Jun 12, 2009 #16
    I can use the density of steel to see how much steel is present, then this is also equal to the weight of displaced mercury. CORRECT?

    Then see how deep it is using m=Vp. CORRECT?
     
  18. Jun 12, 2009 #17

    Redbelly98

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